Question

\mathrm{~A}\( voltaic cell is based on the reaction $$ \mathrm{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if \)75.0 \mathrm{~g}\( of \)\mathrm{Sn}$ is consumed?

Short answer

The maximum electrical work that the cell can accomplish when 75.0 g of Sn is consumed is \( 82,888.7 \mathrm{~Joules} \).

Step-by-step solution

Write down the balanced redox reaction

In this exercise, the balanced redox reaction is given as: $$ \mathrm{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$

Calculate moles of Sn consumed

We are given that 75.0 g of Sn is consumed. To find out the moles of Sn consumed, use the molar mass of Sn: Molar mass of Sn (Tin) = 118.71 g/mol Moles of Sn consumed = \( \frac{75.0 \thinspace g}{118.71 \thinspace g/mol} \) Moles of Sn consumed = 0.6316 moles

Calculate the cell potential

Under standard conditions, we can find the cell potential using the standard reduction potentials of both half-reactions from the given reaction. The reduction potentials can be found in any standard reference material. For Sn half-reaction: $$ \mathrm{Sn^{2+}}(a\thinspace q) + 2 e^{-} \longrightarrow \mathrm{Sn}(s) \quad E^{\circ} = -0.14 \mathrm{V} $$ For I2 half-reaction: $$ \mathrm{I}_{2}(s) + 2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a \thinspace q) \quad E^{\circ} = 0.54 \mathrm{~V} $$ Now, we need to calculate the overall cell potential (E°cell) using the two half-reaction potentials: E°cell = E°(cathode) - E°(anode) E°cell = 0.54V - (-0.14V) E°cell = 0.68V

Calculate the maximum electrical work in Joules

Now, we need to calculate the maximum electrical work that can be accomplished by the cell. We will use the following formula: W = -nFE°cell Where: W = Maximum electrical work n = Moles of electrons F = Faraday's constant (96,485 C/mol) E°cell = Cell potential From the balanced redox reaction, it is clear that 2 moles of electrons are transferred per mole of Sn consumed. Hence, the total moles of electrons (n) can be calculated as: n = 2 × 0.6316 moles n = 1.2632 moles Now, we can plug in the values into the formula for maximum electrical work: W = -(1.2632 moles) × (96,485 C/mol) × (0.68V) W = -82,888.7 J The negative sign indicates that work is done by the cell (as it is a voltaic cell). Therefore, the maximum electrical work the cell can accomplish is 82,888.7 Joules.

Key concepts

Redox Reaction

A redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons between two substances. In a voltaic cell, these reactions are the cornerstone because they enable the conversion of chemical energy into electrical energy. In the given exercise, the redox reaction can be expressed by the equation: \( \mathrm{Sn}(s) + \mathrm{I}_{2}(s) \rightarrow \mathrm{Sn}^{2+}(aq)+2 \mathrm{I}^{-}(aq) \). Here, tin (Sn) loses electrons, undergoing oxidation, whereas iodine \( (\mathrm{I}_2) \) gains the electrons, resulting in reduction. These half-reactions are crucial:

• Oxidation (loses electrons, thus is the anode): \(\mathrm{Sn}(s) \rightarrow \mathrm{Sn}^{2+}(aq) + 2 e^{-}\)
• Reduction (gains electrons, thus is the cathode): \(\mathrm{I}_2(s) + 2 e^{-} \rightarrow 2 \mathrm{I}^{-}(aq)\)

In a voltaic cell, electrons are naturally transferred from the substance being oxidized to the one being reduced. This electron flow can then be used to do work, such as lighting a bulb or running a motor, marrying chemistry with practical electricity applications.

Cell Potential

Cell potential (also known as electromotive force, EMF) is a measure of the voltage or electric potential difference between two half-cells in a voltaic cell. It's imperative because it indicates how much driving force the redox reaction has to push electrons through the circuit.To calculate the cell potential for the reaction \(\mathrm{Sn}(s) + \mathrm{I}_2(s) \rightarrow \mathrm{Sn}^{2+}(aq) + 2 \mathrm{I}^{-}(aq)\), we consider the standard reduction potentials of the half-reactions:

• Sn half-reaction: \( E^{\circ} = -0.14 \mathrm{V} \)
• I2 half-reaction: \( E^{\circ} = 0.54 \mathrm{V} \)

The overall cell potential can be found using the equation:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]Which simplifies to:\[ E^{\circ}_{\text{cell}} = 0.54 \mathrm{V} - (-0.14 \mathrm{V}) = 0.68 \mathrm{V} \]This positive cell potential means that the reaction is spontaneous, further affirming the flow of electrons in the redox reaction, making the voltaic cell functional for generating electrical energy.

Faraday's Constant

Faraday's constant is an essential concept in the study of electrochemistry. It represents the charge of one mole of electrons, calculated to be approximately 96,485 Coulombs per mole (C/mol).This constant is often used in calculations to determine how much work can be done by a voltaic cell. When you understand how electrons transfer during a redox reaction, Faraday's constant helps you connect that electron transfer to practical, measurable electric charge. In the exercise, using the formula \( W = - nFE^{\circ}_{\text{cell}} \), where:

• \( n = 1.2632 \) moles of electrons
• \( F = 96,485 \mathrm{C/mol} \)
• \( E^{\circ}_{\text{cell}} = 0.68 \mathrm{V} \)

...we calculate maximum electrical work. This formula directly informs us of the amount of energy, in joules, that can be harnessed from the chemical reaction occurring in the voltaic cell. Thus, Faraday's constant becomes a bridge between chemical reaction equations and quantifying electrical output, making it indispensable in predicting and understanding the capabilities of electrochemical cells.